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CHAPTER 7. BINARY DATA COMMUNICATION
Problem 7.5 Use the z|required values found in the solution of Problem 7.2 along with z|required = A2 /N0 R to get R = A2 / z|required N0 . Substitute N0 = 10−6 V2 /Hz, A = 20 mV to get R = 400/ z|required . This results in the values given in the table below. PE z|required , dB (ratio) R, bps 10−4 10−5 10−6
8.39 (6.9) 9.58 (9.08) 10.53 (11.3)
57.97 43.05 35.4
Problem 7.6 The integrate-and-dump detector integrates over the interval [0, T − δ] so that the SNR is z0 =
A2 (T − δ) N0
instead of z = A2 T /N0 . Thus T z0 =1− z δ and the degradation in z in dB is µ ¶ T D = −10 log10 1 − δ Using δ = 10−6 s and the data rates given, we obtain the following values of D: (a) for 1/T = R = 10 kbps, D = 0.04 dB; (b) for R = 50 kbps, D = 0.22 dB; (c) for R = 100 kbps, D = 0.46 dB. Problem 7.7 (a) For the pulse sequences (−A, −A) and (A, A), which occur half´the time, no degradation ³p 2Eb /N0 . The error probability results from timing error, so the error probability is Q for the sequences (−A, hp A) and (A, −A), whichi occur the other half the time, result in the 2Eb /N0 (1 − 2 |∆T | /T ) (sketches of the pulse sequences with the error probability Q integration interval offset from the transition between pulses is helpful here). Thus, the average error probability is the given result in the problem statement. Plots are shown in Figure 7.1 for the giving timing errors. (b) The plots given in Figure 7.1 for timing errors of 0 and 0.15 indicate that the degradation at PE = 10−6 is about 2.8 dB.